如何找到只有 0 和 1 除以给定数字的最小数字? [英] How to find the smallest number with just 0 and 1 which is divided by a given number?

问题描述

每个正整数除以表示(以 10 为底)仅包含零和一的数字.

Every positive integer divide some number whose representation (base 10) contains only zeroes and ones.

可以证明:

考虑数字 1、11、111、1111 等，直到 111...1，其中最后一个数字有 n+1 位数字.称这些数字为 m₁、m₂、...、m_n+1.每个都有一个除以 n 时的余数，且其中两个余数必须相同.因为它们有 n+1 个，但余数只能取 n 个值.这是著名而有用的鸽巢原理"的应用；

Consider the numbers 1, 11, 111, 1111, etc. up to 111... 1, where the last number has n+1 digits. Call these numbers m₁, m₂, ... , m_n+1. Each has a remainder when divided by n, and two of these remainders must be the same. Because there are n+1 of them but only n values a remainder can take. This is an application of the famous and useful "pigeonhole principle";

假设余数相同的两个数是m_i和m_j, 与我 <j.现在从较大的减去较小的.结果数 m_i−m_j 由 j - i 个 1 和 i 个零组成，必须是 n 的倍数.

Suppose the two numbers with the same remainder are m_i and m_j , with i < j. Now subtract the smaller from the larger. The resulting number, m_i−m_j, consisting of j - i ones followed by i zeroes, must be a multiple of n.

但是如何找到最小的答案呢?并且有效?

But how to find the smallest answer? and effciently?

推荐答案

好问题.我使用 BFS 通过中间相遇和其他一些修剪来解决这个问题.现在我的代码可以在合理的时间内解决 n<10⁹.

Nice question. I use BFS to solve this question with meet-in-the-middle and some other prunings. Now my code can solve n<10⁹ in a reasonable time.

#include <cstdio>
#include <cstring>

class BIT {
private: int x[40000000];
public:
    void clear() {memset(x, 0, sizeof(x));}
    void setz(int p, int z) {x[p>>5]=z?(x[p>>5]|(1<<(p&31))):(x[p>>5]&~(1<<(p&31)));}
    int bit(int p) {return x[p>>5]>>(p&31)&1;}
} bp, bq;

class UNIT {
private: int x[3];
public: int len, sum;
    void setz(int z) {x[len>>5]=z?(x[len>>5]|(1<<(len&31))):(x[len>>5]&~(1<<(len&31)));}
    int bit(int p) {return x[p>>5]>>(p&31)&1;}
} u;

class MYQUEUE {
private: UNIT x[5000000]; int h, t;
public:
    void clear() {h = t = 0;}
    bool empty() {return h == t;}
    UNIT front() {return x[h];}
    void pop() {h = (h + 1) % 5000000;}
    void push(UNIT tp) {x[t] = tp; t = (t + 1) % 5000000;}
} p, q;

int n, md[100];

void bfs()
{
    for (int i = 0, tp = 1; i < 200; i++) tp = 10LL * (md[i] = tp) % n;

    u.len = -1; u.sum = 0; q.clear(); q.push(u); bq.clear();
    while (1)
    {
        u = q.front(); if (u.len >= 40) break; q.pop();
        u.len++; u.setz(0); q.push(u);
        u.setz(1); u.sum = (u.sum + md[u.len]) % n;
        if (!bq.bit(u.sum)) {bq.setz(u.sum, 1); q.push(u);}
        if (!u.sum) {
            for (int k = u.len; k >= 0; k--) printf("%d", u.bit(k));
            puts(""); return;
        }
    }

    u.len = 40; u.sum = 0; p.clear(); p.push(u); bp.clear();
    while (1)
    {
        u = p.front(); p.pop();
        u.len++; u.setz(0); p.push(u);
        u.setz(1); u.sum = (u.sum + md[u.len]) % n;
        if (!bp.bit(u.sum)) {bp.setz(u.sum, 1); p.push(u);}
        int bf = (n - u.sum) % n;
        if (bq.bit(bf)) {
            for (int k = u.len; k > 40; k--) printf("%d", u.bit(k));
            while (!q.empty())
            {
                u = q.front(); if (u.sum == bf) break; q.pop();
            }
            for (int k = 40; k >= 0; k--) printf("%d", u.bit(k));
            puts(""); return;
        }
    }
}

int main(void)
{
    // 0 < n < 10^9
    while (~scanf("%d", &n)) bfs();
    return 0;
}

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